Show that the ratio of velocities of equal masses in an inelastic collision when one of masses is stationary is `v_(1)/v_(2)=(1-e)/(1+e)`

Explanation

`e=("velocity of speration" ("after collision"))/("velocity of approach" ("before collision"))`
`=((v_(2)-v_(1)))/((u_(1)-u_(2)))=((v_(2)-v_(1)))/((u_(1)-0))=(v_(2)-v_(1))/u_(1)`
implies `v_(2)-v_(1)=eu_(1)`
From the law of conservation of linear momentum,
`mu_(1)=mv_(1)+mv_(2)impliesu_(1)=v_(2)`
Using the equation (2) for `u_(1) `in (1), we get
`v_(2)-v_(1)=e(v_91)+v_(2)`
On simplification, we get
`v_(1)/v_(2)=(1-e)/(1+e)`